Đáp án:
\(\begin{array}{l}
a,\\
P = \dfrac{{ - 7\sqrt 3 + 3}}{3}\\
b,\\
Q = \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c,\\
\left\{ \begin{array}{l}
0 \le x < 25\\
x \ne 4
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
DKXD:\,\,\,\,\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
a,\\
x = 7 - 4\sqrt 3 = 4 - 4\sqrt 3 + 3 = {2^2} - 2.2.\sqrt 3 + {\sqrt 3 ^2} = {\left( {2 - \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} = \left| {2 - \sqrt 3 } \right| = 2 - \sqrt 3 \\
\Rightarrow P = \dfrac{{\sqrt x + 5}}{{\sqrt x - 2}} = \dfrac{{\left( {2 - \sqrt 3 } \right) + 5}}{{\left( {2 - \sqrt 3 } \right) - 2}} = \dfrac{{7 - \sqrt 3 }}{{ - \sqrt 3 }}\\
= \dfrac{{ - \sqrt 3 .\left( {7 - \sqrt 3 } \right)}}{{{{\left( { - \sqrt 3 } \right)}^2}}} = \dfrac{{ - 7\sqrt 3 + 3}}{3}\\
b,\\
Q = \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} - \dfrac{{5\sqrt x - 2}}{{4 - x}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} + \dfrac{{5\sqrt x - 2}}{{x - 4}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 2}} + \dfrac{{5\sqrt x - 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}\\
= \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 2} \right) + 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\left( {x - 3\sqrt x + 2} \right) + 5\sqrt x - 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{x + 2\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}\\
= \dfrac{{\sqrt x }}{{\sqrt x - 2}}\\
c,\\
M = \dfrac{Q}{P} = \dfrac{{\sqrt x }}{{\sqrt x - 2}}:\dfrac{{\sqrt x + 5}}{{\sqrt x - 2}} = \dfrac{{\sqrt x }}{{\sqrt x - 2}}.\dfrac{{\sqrt x - 2}}{{\sqrt x + 5}} = \dfrac{{\sqrt x }}{{\sqrt x + 5}}\\
M < \dfrac{1}{2}\\
\Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x + 5}} < \dfrac{1}{2}\\
\Leftrightarrow 2\sqrt x < \sqrt x + 5\\
\Leftrightarrow \sqrt x < 5\\
\Leftrightarrow x < 25\\
\Rightarrow \left\{ \begin{array}{l}
0 \le x < 25\\
x \ne 4
\end{array} \right.
\end{array}\)
